Question

The scores on an exam are normally distributed, with a mean of 74 and a standard deviation of 7. What percent of the scores are less than 81?

Answer 1

Mean = 74
Standard deviation = 7

For 81%, the Z-score is
Z=(X-mean)/(standard deviation)
=(81-74)/7
=1

So look up table of normal distribution for
P(Z<1)=0.8413
=>
On average, 84% of scores are less than 81.